binary 1

[benbotektips]

Any way to convert a number field to its binary equivalent (ex. BINARY(15) = 1111, BINARY(13) = 1101 )?

 

[SantaMufasa]

Benbo,

Further below is my script that converts any valid integer in any base to its equivalent in any other base. For example, using your data:

col a heading "Base|Conversions" format a11
select baseconv(1111,2,10) a from dual;

Base
Conversions
-----------
15

select baseconv(1101,2,10) a from dual;

Base
Conversions
-----------
13

select baseconv('1234567890abcdefghijklmnopqrstuvwxyz',36,16) from dual;

BASECONV('1234567890ABCDEFGHIJKLMNOPQRSTUVWXYZ',36,16)
------------------------------------------------------
20A45C4A82A3EB406AF773B5C3F6BBB6ABA606A213EA58

The code you need to run to create the "baseconv" function is:
[/code]
REM **************************************************************
REM David L. Hunt (file author) distributes this and other
REM files/scripts for educational purposes only, to illustrate the
REM use or application of various computing techniques. Neither the
REM author nor "Dasages, LLC" makes any warranty regarding this
REM script's fitness for any industrial application or purpose nor is
REM there any claim that this or any similarly-distributed scripts
REM are error free or should be used for any purpose other than
REM illustration.
REM
REM Please contact the author via email (dave@dasages.com) when
REM you have comments, suggestions, and/or difficulties with this
REM package or its functions.
REM
REM [Please keep the above disclaimer and the embedded electronic
REM documentation with this script.]
REM **************************************************************
REM About this script/file:
REM
REM NAME: BASECONV.SQL - PL/SQL code to translate positive or
REM negative integer values from any base (radix)
REM (base 36 max.) to its counterpart value in any other
REM base (base 36 max.).
REM
REM AUTHOR: Dave Hunt
REM Co-principal, Dasages, LLC
REM 1-801-733-5333
REM
REM **************************************************************
REM Usage:
REM BASECONV (<value>,<source-base>,<target-base>)
REM
REM ...where <value> is some "number" using valid "numerals" in
REM the <source-base>. For example, in base 2, valid numerals are
REM "0" and "1"; in base 36, valid numerals are "0" thru "z".
REM When using bases 11 thru 36, you specify alpha characters for
REM numerals beyond decimal '9'. Alpha characters may be in either
REM upper or lower case.
REM
REM ...where <source-base> is any base (radix) between 2 and 36.
REM ...where <target-base> is any base (radix) between 2 and 36.
REM Be sure to specify both source and target bases in base 10.
REM
REM Sample invocations:
REM
REM col x heading "Conversion" format a20
REM select baseconv(255,10,2) x from dual;
REM Conversion
REM ----------
REM 11111111
REM
REM select baseconv('Zebra',36,10) x from dual;
REM Conversion
REM ----------
REM 59454982
REM
REM select baseconv('aaaaaaaaaaaaaaaa',16,2) x from dual;
REM Conversion
REM ----------------------------------------------------------------
REM 1010101010101010101010101010101010101010101010101010101010101010
REM **************************************************************
Create or replace function BaseConv
( ValueIn in varchar2 -- incoming value to convert
, RadFrom in number -- source base
, RadOut in number -- target base
)
return varchar2 -- outgoing value in target base
is
ValIn varchar2(1000);
Sign char;
LenIn number;
Base10Value number;
DigitPool varchar2(50) := '123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
DigitHold varchar(1);
HighPower number;
CurrValue number;
CurrDigit number;
ResultingValue varchar(2000);
function GetDigit10 (InDigit in varchar2, RadIn in number) return number
is
bad_digit exception;
pragma exception_init(bad_digit,-6502);
begin
if InDigit = '0' then
return 0;
end if;
DigitHold := upper(InDigit);
for i in 1..RadIn-1 loop
if DigitHold = substr(DigitPool,i,1) then
return i;
end if;
end loop;
raise_application_error(-20000,'Illegal digit, "'||InDigit||'" for base "'||RadIn||'"');
end;
begin
ValIn := ValueIn;
if substr(ValIn,1,1) = '-' then
Sign := '-';
ValIn := substr(ValIn,2);
else
Sign := null;
end if;
LenIn := length(nvl(ValIn,'0'));
Base10Value := 0;
for i in 1..LenIn loop
Base10Value := Base10Value +
GetDigit10(substr(ValIn,i,1),RadFrom) * power(RadFrom,LenIn-i);
end loop;
for i in 1..1000 loop
if power(RadOut,i) > Base10Value then
HighPower := i-1;
exit;
end if;
end loop;
CurrValue := Base10Value;
ResultingValue := null;
for i in 0..HighPower loop
CurrDigit := floor(Currvalue / power(RadOut,HighPower-i));
CurrValue := Currvalue - (CurrDigit * power(RadOut,HighPower-i));
if CurrDigit = 0 then
ResultingValue := ResultingValue||'0';
else
ResultingValue := ResultingValue||substr(DigitPool,CurrDigit,1);
end if;
end loop;
return sign||ResultingValue;
end;
/
[/code]

Mufasa
(aka Dave of Sandy, Utah, USA)
e-mail: dave@dasages.com
@ 01:49 (16Nov04) UTC (aka "GMT" and "Zulu"),
@ 18:49 (15Nov04) Mountain Time

 

[benbotektips]

Thanx, Santa. This is a simple, practical solution.

 

[benbotektips]

An alternative is available, in Oracle and binary anyway. There's a function 'bitand (number1, number2)', which ANDs the (base2) bits. So, bitand (status, 1) will be true u if the 1st bit is on, bitand (status, 128), will be true u if the 7th bit is on, etc.